14) E

In order for a species to disproportionate, the species must occur twice in the products.

In reactions 1 and 4, all species occur only once in the products.

In reaction 2,
Cu appears twice in the products.
Oxidation state of Cu in Cu₂O in the reactants = +1
Oxidation state of Cu in Cu₂ in the products = 0 (so reduced)
Oxidation state of Cu in CuO in the products = +2 (so oxidised)

In reaction 3,
Cl appears twice in the products.
Oxidation state of Cl in Cl₂ in the reactants = 0
Oxidation state of Cl in HCl in the products = -1 (so reduced)
Oxidation state of Cl in HClO in the products = +1 (so oxidised)

In reaction 5,
Hg appears twice in the products.
Oxidation state of Hg in Hg₂Cl₂ in the reactants = +1
Oxidation state of Hg in Hg in the products = 0 (so reduced)
Oxidation state of Hg in HgCl₂ in the products = +2 (so oxidised)

In reactions 2, 3 and 5, simultaneous oxidation and reduction of the same species has occurred.